# different angles same solution

• Jan 18th 2009, 03:58 PM
Serialkisser
different angles same solution
While a student was playing with their calculator they found that sometimes the answers produced from taking the sine and cosine of different angles were the same answer. The information below are some of the examples when this occurs.

sin 55* = 0.819152044 and cos 35* = 0.819152044
sin 30* = 0.5 and cos 60* = 0.5
sin 15* = 0.258819045 and cos 75* = 0.258819045

I have to explain why these calculations produce the same answer.

I looked numerous times at the formula,but i cant come up with anything.
• Jan 18th 2009, 04:12 PM
Mush
Quote:

Originally Posted by Serialkisser
While a student was playing with their calculator they found that sometimes the answers produced from taking the sine and cosine of different angles were the same answer. The information below are some of the examples when this occurs.

sin 55* = 0.819152044 and cos 35* = 0.819152044
sin 30* = 0.5 and cos 60* = 0.5
sin 15* = 0.258819045 and cos 75* = 0.258819045

I have to explain why these calculations produce the same answer.

I looked numerous times at the formula,but i cant come up with anything.

In the function $y=\sin(x)$, the function is defined for ALL values of $x$. I.e. $x \in (-\infty,\infty)$

So now we know how $x$ varies, what about $y$ ?

The $y$ values in $y = sin(x)$ range between the amplitude of the sine wave. That amplitude is $1$ . For each and ever real number between $-1$ and $1$ inclusive, there is a $y$ value for this function! i.e. $y \in [-1,1]$.

In the function $y=\cos(x)$, the function is defined for ALL values of $x$. I.e. $x \in (-\infty,\infty)$
So now we know how $x$ varies, what about $y$ ?
The $y$ values in $y = cos(x)$ range between the amplitude of the cosine wave. That amplitude is $1$ . For each and ever real number between $-1$ and $1$ inclusive, there is a $y$ value for this function! i.e. $y \in [-1,1]$ .
In conclusion, both $\sin(x)$ and $\cos(x)$ are defined on the SAME $y$ intervals and the same $x$ intervals, and each of them has a value at every point on this intervals. Hence, for every $y$ such that $y = \sin(\alpha)$, there is also $y = \cos(\beta)$.
$cos(\pm x) = cos(x) = sin(\pm x + 90 \text{ degrees})$